How is $\frac{3^n}{5^{n-1}}$ a geometric series?

Question

How is $$\frac{3^n}{5^{n-1}}$$

a geometric series? I don't understand how the geometric series test can be applied to it to determine convergence .

Answer 1

$\frac{3^n}{5^{n-1}} = 5(\frac{3^n}{5^n}) =5(\frac{3}{5})^n$

Answer 2

It is the same as $3*\left(\frac{3}{5}\right)^{n-1}$

or

$\frac{3^n}{5^{n-1}} = \frac{3^n}{5^{-1}*5^{n}} = 5*\frac{3^n}{5^n} = 5*\left(\frac35\right)^n$

Answer 3

$\frac{3^n}{5^{n-1}} = 5\cdot\frac{3^n}{5^n} = 5\cdot\left(\frac35\right)^n$

Answer 4

A geometric sequence is a sequence of the form $$\begin{cases} g_{0} &\\ g_{n} &= g_{0}q^{n} \end{cases}$$ for $q\in\mathbb{R}^{+}_{0}$.

Here, by rewriting $$\frac{3^n}{5^{n-1}}=5\frac{3^{n}}{5^{n}}=5\left(\frac{3}{5}\right)^{n}$$ you can see it is a geometric sequence with $g_{0}=5$ and $q=\frac{3}{5}$

Answer 5

Remember that $$3^n = 3^{n-1} \cdot 3$$

So $$\frac{3^n}{5^{n-1}} = \frac{3^{n-1} \cdot 3}{5^{n-1}}$$

But this is the same as

$$3 \cdot \frac{3^{n-1}}{5^{n-1}} =3 \cdot \left(\frac{3}{5}\right)^{n-1} $$

This is a geometric series with common ratio $3/5$.